2 Prerequisites

The polygamma function, $\psi _{k}(z)$ , is a meromorphic function on ${\mathbb C}$ defined by

$$ \begin{align*} \psi_{k}(z)=\frac{d^{k+1}}{dz^{k+1}}\log\Gamma(z) = (-1)^{k+1}k!\sum_{n=0}^{\infty}\frac{1}{(n+z)^{k+1}}, \quad z\ne 0, -1, -2, \ldots. \end{align*} $$

If m is a natural number, then

$$ \begin{align*} \psi_{k}(m)=\psi_{k}(1)+(-1)^{k+2}k!\sum_{n=1}^{m-1}\frac{1}{(n+1)^{k+1}}. \end{align*} $$

The digamma function, $\psi (z) = \psi _0(z) = \Gamma '(z)/\Gamma (z)$ , satisfies the functional equations

$$ \begin{align*} \psi(z+1)=\psi(z)+\frac{1}{z}, \quad \psi(1-z)=\psi(z)+\pi\cot{\pi z}. \end{align*} $$

Consequently, the polygamma function satisfies the functional equations

$$ \begin{align*} \psi_{k}(z+1)=\psi_{k}(z)+\frac{(-1)^{k}k!}{z^{k+1}}, \quad (-1)^{k}\psi_{k}(1-z)=\psi_{k}(z)+\frac{d^{k}}{dz^{k}}({\pi\cot{\pi z}}). \end{align*} $$

Recall that, for z not an integer,

$$ \begin{align*} \pi\cot\pi z=\frac{1}{z}+\sum_{n=1}^{\infty}\bigg(\frac{1}{z-n}+\frac{1}{z+n}\bigg). \end{align*} $$

We end this section by listing some irrationality results relevant to our work. In $1978$ , Apéry astonished the mathematics community by proving the irrationality of the function $\zeta (3)$ . Following that, Ball and Rivoal [Reference Ball and Rivoal2, Reference Rivoal12] made the next remarkable breakthrough.

Theorem 2.1 (Ball and Rivoal, [Reference Ball and Rivoal2]).

Let $\epsilon>0$ . For any $s\geq 3$ odd and sufficiently large with respect to $\epsilon $ ,

$$ \begin{align*} \dim_{\mathbb{Q}} \mathrm{span}_{\mathbb{Q}}\{1,\zeta(3),\zeta(5),\ldots,\zeta(s)\}\geq \frac{1-\epsilon}{1+\log {2}}\log {s}. \end{align*} $$

Fischler et al. [Reference Fischler, Sprang and Zudilin5] recently proved the following result.

3 Proof of Theorem 1.2

We begin with the odd case. For $z=x+iy$ with $x,y\in \mathbb {R}$ ,

$$ \begin{align*} \psi_{2k+1}(x+iy) & =(2k+1)!\sum_{n=0}^{\infty}\frac{(n+x-iy)^{2k+2}}{((n+x)^2+y^2)^{2k+2}} \\ & =(2k+1)!\sum_{n=0}^{\infty}\frac{\sum_{r=0}^{2k+2}\binom{2k+2}{r}(n+x)^{2k+2-r}(-iy)^{r}}{((n+x)^2+y^2)^{2k+2}}. \end{align*} $$

The real part of $\psi _{2k+1}(x+iy)$ is given by

$$ \begin{align*} &(2k+1)!\bigg[\sum_{n=0}^{\infty}\frac{\sum_{r=0,4|r}^{2k+2}\binom{2k+2}{r}(n+x)^{2k+2-r}(y)^{r}}{((n+x)^2+y^2)^{2k+2}}\bigg] \\ & \quad- (2k+1)!\bigg[\sum_{n=0}^{\infty}\frac{\sum_{r=2, 4|(r-2)}^{2k+2}\binom{2k+2}{r}(n+x)^{2k+2-r}(y)^{r}}{((n+x)^2+y^2)^{2k+2}}\bigg]. \end{align*} $$

When z is a real number (which is not a pole), that is, when $y=0$ ,

$$ \begin{align*} \operatorname{Re} \psi_{2k+1}(x) = (2k+1)!\sum_{n=0}^{\infty}\frac{1}{(n+x)^{2k+2}}> 0. \end{align*} $$

Consequently, $\psi _{2k+1}(x)\neq 0$ and so $\psi _{2k+1}$ has no real zeros.

We now turn to the more involved case when k is even. We see that $\psi _{2k}(x)$ has no real zeros with $x\geq 0$ . However, unlike the odd case, $\psi _{2k}$ does have real zeros. We divide our investigation into two cases.

Case 1: $\psi _{2k}$ has no real zeros in the intervals $(-m, -m + 1/2]$ for all integers $m \ge 1$ . The derivative of $\psi _{2k}$ is given by

$$ \begin{align*} \psi_{2k+1}(x)=(2k+1)!\sum_{n=0}^{\infty}\frac{1}{(n+x)^{2k+2}}, \end{align*} $$

which is positive. Thus, the function $\psi _{2k}$ is a strictly increasing function of x in each of the intervals $I_{n}= (-n,-n+1)$ for $n=1,2,3,\ldots. $ Appealing to the functional equation, we show that $ \psi _{2k}(-m+{1}/{2})<0$ . This, along with the strict monotonicity noted above, shows that $ \psi _{2k}$ has no zero in $(-m, -m + 1/2]$ for all integers $m \ge 1$ .

To this end, we observe that

$$ \begin{align*} &\psi_{2k}\bigg( -m+\frac{1}{2}\bigg)=-2k!\sum_{n=0}^{\infty}\frac{1}{(n-m+{1}/{2})^{2k+1}}=-2k!~2^{2k+1}\sum_{n=0}^{\infty}\frac{1}{(2(n-m)+1)^{2k+1}} \\ &\quad= -2k!~2^{2k+1} \bigg[\bigg(\sum_{n=0}^{m-1} + \sum_{n=m}^{2m-1} + \sum_{n=2m}^\infty\bigg) \frac{1}{(2(n-m)+1)^{2k+1}}\bigg]. \end{align*} $$

The terms in the first two sums cancel in pairs and we are left with

$$ \begin{align*} \psi_{2k}\bigg( -m+\frac{1}{2}\bigg)=2k!~2^{2k+1}\bigg[-\frac{1}{(2m+1)^{2k+1}}-\frac{1}{(2m+3)^{2k+1}}-\cdots \bigg]<0, \end{align*} $$

which proves the claim.

Case 2: Real zeros in intervals of the form $(-m + 1/2, -m +1)$ . Consider

$$ \begin{align*}\frac{-p}{q}=-m+\frac{a}{q}, \,\,\,\,\, \frac{q}{2}< a\leq q-1,\end{align*} $$

where the denominator q is an odd positive integer. We show that

$$ \begin{align*} \psi_{2k}\bigg(-m+\frac{q+1}{2q}\bigg)>0 \end{align*} $$

for $q\leq 228$ . This, along with the fact that $\psi _{2k}$ is strictly increasing in each interval $(-m , 1 -m)$ , ensures that

$$ \begin{align*} \psi_{2k}\bigg(-m+\frac{a}{q}\bigg)\geq \psi_{2k}\bigg(-m+\frac{114}{227}\bigg)>0. \end{align*} $$

Since $2qm-q-1<2mq-q+1 $ ,

$$ \begin{align*} & \frac{\psi( -m+{q+1}/{2q})}{ 2k!(2q)^{2k+1}} \\ &\quad \geq \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{1}{(2qm+q+1)^{2k+1}}-\frac{1}{(2mq+3q+1)^{2k+1}}-\cdots\bigg] \\ &\quad \geq \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{1}{(2qm)^{2k+1}}-\frac{1}{(2q(m+1))^{2k+1}}-\cdots\bigg] \\ &\quad> \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{1}{(2q)^{2k+1}}\bigg(\zeta(2k+1)-\sum_{n=1}^{m-1}\frac{1}{n^{2k+1}}\bigg)\bigg] \\ &\quad \geq \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{1}{(2q)^{2k+1}}(\zeta(2k+1)-1)\bigg] \\ &\quad \geq \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{0.21}{(2q)^{2k+1}}\bigg] \quad(\mbox{since } \zeta(2k+1)\leq 1.21 \mbox { for all } k\in \mathbb{N}) \\ &\quad\geq \bigg[\frac{1}{(q-1)^{2k+1}}-\frac{1}{(q+1)^{2k+1}}-\frac{0.02625}{q^{2k+1}}\bigg]. \end{align*} $$

We claim that the last quantity is positive for $q \leq 228$ and all k. Consider

$$ \begin{align*}f(x)=\frac{1}{(x-1)^{m}}-\frac{1}{(x+1)^{m}}-\frac{0.02625}{x^{m}}.\end{align*} $$

Now, $f(x)>0$ is equivalent to

$$ \begin{align*}1>\bigg(1-\frac{2}{x+1}\bigg)^{m}+0.02625\bigg(1-\frac{1}{x}\bigg)^{m}.\end{align*} $$

The last inequality holds for $x=228$ and $m=3$ and the function on the right-hand side is an increasing function of x, so it holds for all $x \leq 228$ . For this range of x, the function on the right-hand side is a decreasing function of m for $m\geq 3$ . This implies that $f(x)>0$ for $x\leq 228$ and $m\geq 3$ . Since q is an odd integer,

$$ \begin{align*}\psi_{2k}\big( -m+\tfrac{114}{227}\big)>0.\end{align*} $$

So the real zeros of $\psi _{2k}$ will lie in the intervals $(-m+1/2,-m+114/227)$ for all $m>0$ .

Next, we show that $\psi _{2k}$ will have a zero in the interval $(-n,-n+1)$ for all ${n=1,2,3,\ldots. }$ In the preceding calculations, we saw that

$$ \begin{align*} \psi_{2k}\bigg(\frac{-p}{q}\bigg)< 0, \quad\mbox{where } \frac{-p}{q}=-m+\frac{a}{q}, a\leq\frac{q}{2} \end{align*} $$

and

$$ \begin{align*} \psi_{2k}\bigg( -m+\frac{a}{q}\bigg)\geq \psi_{2k}\bigg( -m+\frac{114}{227}\bigg)>0. \end{align*} $$

Since $\psi _{2k}(x)$ is continuous and strictly increasing in each interval $I_{n}= (-n,-n+1)$ for $n=1,2,3,\ldots ,$ it follows that $\psi _{2k}(x)$ has exactly one zero in each of these intervals.

Remark 3.1. We have proved that $\psi _{2k}$ has one real zero, say, $x_m$ , in the interval ${(-m+1/2,-m+114/227)}$ . Write $x_{m}=-m+y_{m}$ , where $y_{m}\in (0,1)$ .

We claim that the sequence $\{y_{m}\}$ is strictly decreasing and that $\psi _{2k}( -m+{1}/{2}) \to 0$ as $m \to \infty $ .

Proof. Suppose that $x_{m}=-m+y_{m}$ and $x_{m'}=-m'+y_{m'}$ are both zeros of $\psi _{2k}$ , where $m>m'$ and $y_{m}\ge y_{m'}$ . From the functional equation,

$$ \begin{align*} 0 & =\psi_{2k}(-m+y_{m})-\psi_{2k}(-m'+y_{m'}) \\ & =\psi_{2k}(-m+y_{m})-\psi_{2k}(-m+y_{m'})-\sum_{n=m'+1}^{m}\frac{2k!}{(-n+y_{m'})^{2k+1}} \\ & =\psi_{2k}(-m+y_{m})-\psi_{2k}(-m+y_{m'})+\sum_{n=m'+1}^{m}\frac{2k!}{(n-y_{m'})^{2k+1}}>0 \end{align*} $$

because $\psi _{2k}$ is strictly increasing in $(-m,-m+1)$ . This is a contradiction, so $y_m<y_{m'}$ .

Now, we come to the second assertion,

$$ \begin{align*} \psi_{2k}\bigg( -m+\frac{1}{2}\bigg) & =-2k!\sum_{n=0}^{\infty}\frac{1}{(n-m+{1}/{2})^{2k+1}}=-2k!~2^{2k+1}\sum_{n=0}^{\infty}\frac{1}{(2(n-m)+1)^{2k+1}} \\ & =2k!~2^{2k+1}\bigg[\sum_{n=0}^{m}\frac{1}{(2n+1)^{2k+1}}-\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2k+1}}\bigg] \to 0 \end{align*} $$

as $m \rightarrow \infty $

4 Proof of Theorem 1.3

We begin with a proof of Theorem 3(b). For $x\ge 0$ ,

$$ \begin{align*} \psi_{1}(x+iy) & =\sum_{n=0}^{\infty}\frac{1}{(n+x+iy)^{2}}=\sum_{n=0}^{\infty}\frac{(n+x-iy)^{2}}{((n+x)^{2}+y^{2})^{2}} \\ & =\sum_{n=0}^{\infty}\frac{(n+x)^{2}-y^{2}}{((n+x)^{2}+y^{2})^{2}}-2iy\sum_{n=0}^{\infty}\frac{n+x}{((n+x)^{2}+y^{2})^{2}}. \end{align*} $$

Now, $\operatorname{Im} \psi _{1}(x+iy)$ will be zero if and only if either

$$ \begin{align*} y=0 \quad\mbox{or}\quad\sum_{n=0}^{\infty}\frac{n+x}{((n+x)^{2}+y^{2})^{2}}=0. \end{align*} $$

If $y=0$ , then by Theorem 1.2, $\operatorname{Re} \psi _{1}(x+iy)\neq 0$ , so $\psi _{1}(x+iy)\not =0$ . If $y\ne 0$ , then

$$ \begin{align*}\sum_{n=0}^{\infty}\frac{n+x}{((n+x)^{2}+y^{2})^{2}}>0\end{align*} $$

for $x\geq 0$ and, again, $\operatorname{Im} \psi _{1}(x+iy)\not =0$ .

Now, we turn to part (a) of the theorem. It is enough to show that $\psi _1(-m+iy) \neq 0$ , where $m\in \mathbb {N}, y\in \mathbb {R}$ . As before,

$$ \begin{align*}\operatorname{Im}\psi_{1}(-m+iy)=-2y\sum_{n=0}^{\infty}\frac{n-m}{((n-m)^{2}+y^{2})^{2}}.\end{align*} $$

Hence, $\operatorname{Im} \psi _{1}(-m+iy)$ will be zero if and only if either

$$ \begin{align*}y=0 \quad\mbox{or}\quad \sum_{n=0}^{\infty}\frac{n-m}{((n-m)^{2}+y^{2})^{2}}=0.\end{align*} $$

If $y=0$ , then $\psi _{1}(z)$ has a pole at $z=-m$ , so this cannot be a zero of $\psi _{1}(z)$ . If $y\ne 0$ , then

$$ \begin{align*} \sum_{n=0}^{\infty}\frac{n-m}{((n-m)^{2}+y^{2})^{2}}=\frac{m+1}{((m+1)^{2}+y^{2})^{2}}+\frac{m+2}{((m+2)^{2}+y^{2})^{2}}\cdots>0, \end{align*} $$

because the first $2m$ terms of the series cancel in pairs (for $n=k$ and $n=2m-k$ ), so $\operatorname{Im} \psi _{1}(-m+iy)\ne 0$ . Hence, $\psi _{1}(z)$ has no zero in $\mathbb {Z}+i\mathbb {R}$ . This completes the proof.

5 Proof of Theorem 1.4

Theorem 5.1. For positive integers N and $q>2$ , let $V_{o,N}(q)$ and $V_{e,N}(q)$ denote the ${\mathbb Q}$ -vector spaces

$$ \begin{align*} V_{o,N}(q):= \mathrm{span}_{\mathbb{Q}}\{ \psi_{2k+1} (a/q) : 0\le k \le N , 1 \leq a \leq q , (a,q) =1 \} \end{align*} $$

and

$$ \begin{align*} V_{e,N}(q):= \mathrm{span}_{\mathbb{Q}}\{ 1, \psi_{2k} (a/q) : 1\le k \le N , 1 \leq a \leq q , (a,q) =1 \}.\end{align*} $$

Then $\dim _{\mathbb {Q}}V_{o,N}(q) \gg N$ and $\dim _{\mathbb {Q}}V_{e,N}(q) \gg \log N$ .

Proof. We show that

$$ \begin{align*}\dim_{\mathbb{Q}}\mbox{span}_{\mathbb{Q}}\{\psi_{2k+1}(a/q):0 \leq k\leq N,1\leq a<q,(a,q)=1\}\geq N+1.\end{align*} $$

By the definition of $\psi _{k}(z)$ ,

$$ \begin{align*} \psi_{2k+1}(a/q)=(2k+1)!\sum_{n=0}^{\infty}\frac{1}{(n+a/q)^{2k+2}}=(2k+1)!\zeta(2k+2,a/q). \end{align*} $$

Since

$$ \begin{align*} q^{k}\zeta(k)\prod_{p|q}(1-p^{-k})=\sum_{\substack{a=1\\ (a,q)=1}}^{q-1}\zeta(k,a/q), \end{align*} $$

it follows that $\mbox {span}_{\mathbb {Q}}\{\zeta {(2k+2,a/q)}:0 \leq k\leq N,1\leq a<q,(a,q)=1\}$ contains $\mbox {span}_{\mathbb {Q}}\{\zeta {(2k+2)}:0 \leq k\leq N\}$ . Now, $\dim _{\mathbb {Q}}\mbox {span}_{\mathbb {Q}}\{\zeta {(2k+2)}:0 \leq k\leq N\}= N+1$ , so we get the required result.

Similarly,

$$ \begin{align*} &\dim_{\mathbb{Q}}\mbox{span}_{\mathbb{Q}}\{1,\psi_{2k}(a/q):1 \leq k \leq N, 1\leq a<q,(a,q)=1\} \\ &\quad =\dim_{\mathbb{Q}}\mbox{span}_{\mathbb{Q}}\{1,\zeta{(2k+1,a/q)}:1 \leq k \leq N, 1\leq a<q,(a,q)=1\} \\ &\quad \ge \dim_{\mathbb{Q}}\mbox{span}_{\mathbb{Q}}\{1,\zeta{(2k+1)}:1 \leq k \leq N\}, \end{align*} $$

and the theorem of Ball and Rivoal (Theorem 2.1) gives the desired result.

Theorem 1.4 and the following corollary are immediate consequences of Theorem 5.1.

Corollary 5.2. Let $k,q>1$ be integers. Then the $\mathbb {Q}$ -linear space generated by

$$ \begin{align*}\{\psi_{k}(a/q): k\geq 1, 1\leq a<q,(a,q)=1\}\end{align*} $$

has infinite dimension over $\mathbb {Q}$ .

6 The conjecture of Chowla–Milnor and its ramifications

Gun et al. [Reference Gun, Murty and Rath6] discuss several conjectures on the linear independence of values of the Hurwitz zeta function. The starting point is a question of P. and S. Chowla.

Conjecture 6.1 (Chowla and Chowla, [Reference Chowla and Chowla3]).

Let p be any prime and let f be any rational-valued periodic function with period p. Then

$$ \begin{align*}L(2,f) = \sum_{n=1}^{\infty} \frac{f(n)}{n^2} \neq 0,\end{align*} $$

except for the case when

$$ \begin{align*} f(1)=f(2)=\cdots =f(p-1)=\frac{f(p)}{1-p^{2}}. \end{align*} $$

Milnor interpreted this conjecture in terms of the linear independence of the Hurwitz zeta function and generalised it for all $k>1$ .

Conjecture 6.2 (Milnor, [Reference Milnor9]).

For any integer $k>1$ , the real numbers

$$ \begin{align*} \zeta(k,1/p),\zeta(k,2/p),\ldots,\zeta(k,(p-1)/p) \end{align*} $$

are linearly independent over $\mathbb {Q}$ .

Further, for q not necessarily prime, Milnor suggested the following generalisation of the Chowla conjecture.

Following Gun et al. [Reference Gun, Murty and Rath6], for any integer $k>1$ , define the $\mathbb {Q}$ -linear space $V_{k}(q)$ by

$$ \begin{align*} V_{k}(q)=\mbox{span}_{\mathbb{Q}}\{\zeta(k,a/q): 1\leq a\leq q,(a,q)=1\}. \end{align*} $$

The Chowla–Milnor conjecture asserts that the dimension of $V_{k}(q)$ for $k>1$ is $\phi (q)$ . Gun et al. [Reference Gun, Murty and Rath6] gave the following nontrivial lower bound for this dimension.

They also proposed a stronger version of the Chowla–Milnor conjecture.

We now indicate the relationship between the Chowla–Milnor conjecture and our investigation, which is facilitated by the relationship

$$ \begin{align*} \psi_{k}\bigg(\frac{a}{q}\bigg)=(-1)^{k+1}k!\sum_{n=0}^{\infty}\frac{1}{(n+a/q)^{k+1}}=(-1)^{k+1}k!\zeta(k+1,a/q). \end{align*} $$

Consequently, the Chowla–Milnor conjecture is equivalent to the assertion that, for any $ k> 1$ , the set

$$ \begin{align*} \{ \psi_k(a/q) : 1 \le a \le q, (a,q) = 1\} \end{align*} $$

is linearly independent over ${\mathbb Q}$ .

Proof. Since $\psi _{2k}(p/q)\not =0$ if $p/q>0$ , it enough to consider negative values of p. Suppose that there exist two zeros in this set, say,

$$ \begin{align*} p_{1}/q=-m+a/q \quad \mbox{and} \quad p_{2}/q=-m'+b/q, \quad\mbox{with } (a,q)=(b,q)=1. \end{align*} $$

Note that the functional equation for $\psi _{2k}$ ensures that both $\psi _{2k}(a/q)$ and $\psi _{2k}(b/q)$ are rational numbers. This shows that $\zeta (2k+1,a/q)$ and $\zeta (2k+1,b/q)$ are linearly dependent over $\mathbb {Q}$ , which is in contradiction to the Chowla–Milnor conjecture.

For $k=1$ , the series is $\mathrm {Li}_{1}(z)=-\log {(1-z)}$ , provided that $|z|<1$ .

Analogous to Baker’s theorem on linear forms in logarithms [Reference Baker1], Gun et al. [Reference Gun, Murty and Rath6] proposed the following conjecture for polylogarithms.

We end our paper with the following theorem.

Proof. We derive an extension of Gauss’s formula by an argument of Jensen using roots of unity. This identity is known (see, for example, [Reference Kölbig8]).

We begin with Simpson’s formula. For a power series $f(t)=\sum _{n=0}^{\infty }a_{n}t^{n}$ ,

$$ \begin{align*} \sum_{n=0}^{\infty}a_{qn+m}t^{qn+m}=\frac{1}{q}\sum_{j=0}^{q-1}\omega^{-jm}f(\omega^{j}t), \end{align*} $$

where $\omega $ is a primitive qth root of unity. This follows easily by orthogonality. Now,

$$ \begin{align*} \psi_{k}\bigg(\frac{a}{q}\bigg)=(-1)^{k+1}k!q^{k+1}\sum_{n=0}^{\infty}\frac{1}{(qn+a)^{k+1}}. \end{align*} $$

From the series $\mathrm {Li}_{k}(z)=\sum _{n=1}^{\infty }{z^{n}}/{n^{k}}$ and Simpson’s formula with $\omega =e^{{2\pi i}/{q}}$ and $t=1$ ,

$$ \begin{align*} \psi_{k}\bigg(\frac{a}{q}\bigg)=(-1)^{k+1}k!q^{k}\sum_{j=0}^{q-1}\omega^{-ja}\mathrm{Li}_{k+1}(\omega^{j}). \end{align*} $$

By using this identity and the polylogarithm conjecture, along with

$$ \begin{align*}\psi_{k}\bigg(\frac{a}{q}\bigg)=(-1)^{k+1}k!\sum_{n=0}^{\infty}\frac{1}{(n+a/q)^{k+1}}=(-1)^{k+1}k!\zeta(k+1,a/q)\not=0,\end{align*} $$

we deduce that $\psi _{k}(a/q)$ is transcendental for $1\leq a\leq q$ and $(a,q)=1$ . The first of the two functional equation for $\psi _k(z)$ immediately implies that $\psi _{k}(x)$ is transcendental for any nonintegral rational number x.